Transformation of Covariates

class: center, middle, inverse, title-slide

.title[
# Transformation of Covariates
]
.subtitle[
## Lecture 8
]
.author[
### Manuel Villarreal
]
.date[
### 09/30/24
]

---

### Linear Splines

- In the previous lecture we mention that when we use a linear spline we are 
basically joining two or more independent lines at some point.

- Let's call that point a joint and denote it by the variable `$c_k$` where k 
represents the joint number.

- Formally, for every joint `$c_k$` in the spline (point where two lines join), we 
want to make sure that the following is true:
`$$\lim_{X_i\to c_k^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \lim_{X_i\to c_k^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right)$$`

- This means that we want the expected value of our dependent variable `$Y$` 
to be the same if we approach the joint `$c_k$` from the negative side of 
`$X_i$` that if we approach it from the positive side.

---
### Linear Splines

- In other words, we want our expectations to transition smoothly from one line
to the other without any abrupt jumps.

- Let's check our example from the precious class and see if this condition is
met.

- Formally, our model states that:
`$$\mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(I(X_i > 40)(X_i - 40)\right)$$`

- Where `$I(X_i > 40)$` is an indicator function that takes the value 0 if 
`$X_i\leq 40$` and the value `$1$` otherwise.

- Notice that in this case `$f(X_i) = I(X_i > 40)(X_i - 40)$`.

---
### Linear Splines: continuity

- Now we can check what the limit of our function is.

- Let's start from the side of the negative numbers. Given that the joint is at 
`$c_1 = 40$` we have that:
`$$\lim_{X_i\to c_k^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \lim_{X_i\to c_k^{-}}\left( \beta_0 + \beta_1\left(I(X_i > 40)(X_i - 40)\right) \right)\\ 
\lim_{X_i\to 40^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(\lim_{X_i\to 40^{-}}I(X_i > 40)(X_i - 40)\right)$$`

---
### Linear Splines: continuity

- According to the definition of a limit, we can approach arbitrarily close to 
the value `$c_k$` (in this case `$40$`) from the left, however, we will never reach 
that value. In other words we have that:
`$$\lim_{X_i\to 40^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(0\cdot\lim_{X_i\to 40^{-}}(X_i - 40)\right)$$`

- The limit of our function `$\lim_{X_i\to 40^{-}}(X_i - 40)$` is being 
multiplied by `$0$`. Therefore we know that regardless of the value of `$X_i$`, 
the equation can be simplified as:
`$$\lim_{X_i\to 40^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0$$`

- This means that for people who are `$40$` years old or younger we expect their 
memory performance to be `$\beta_0$`.

---
### Linear Splines: continuity

- Now let's look at the limit from the positive side:
`$$\lim_{X_i\to c_k^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \lim_{X_i\to c_k^{+}}\left( \beta_0 + \beta_1\left(I(X_i > 40)(X_i - 40)\right) \right)\\ 
\lim_{X_i\to 40^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(\lim_{X_i\to 40^{+}}I(X_i > 40)(X_i - 40)\right)$$`

- Now we are approaching the joint `$c_k$` (which is still `$40$`) from the 
positive side of the real line, which means that we can get arbitrarily close 
to 40 without reaching that value. Therefore, we know that `$I(X_i > 40) = 1$`.
`$$\lim_{X_i\to 40^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(\lim_{X_i\to 40^{+}}1\cdot (X_i - 40)\right)$$`

---
### Linear Splines: continuity

- This time there is no zero to multiply the limit, however, we can think about
what will happen as we approach arbitrarily close to `$40$` from the positive 
side.

- We want to figure out: What happens to the `$\lim_{X_i\to 40^{+}}(X_i - 40)$` 
as we move closer and closer to 40? Well, it gets closer and closer to 0!

- In other words, as we move arbitrarily close to 40 the limit will move
arbitrarily close to 0, which means that 
`$$\lim_{X_i\to 40^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 + \beta_1\left(1\cdot 0\right) = \beta_0$$`
--

- And with this we have shown that both expectations are equal!
`$$\lim_{f(X_i)\to c_k^{-}} \mathrm{E}\left(Y_i \mid f(X_i)\right) = \beta_0 = \lim_{f(X_i)\to c_k^{+}} \mathrm{E}\left(Y_i \mid f(X_i)\right)$$`

---
### More than two lines

- What does this mean in practice? Well, it means that we have to make sure that 
two consecutive lines have the same value at their joint.

- Now that we know that we have to make sure that two lines have the same value
at their joint, we can think of adding more lines to a model.

- Let's go back to our average temperature example and model the data from 
the city of Miami using linear splines.

---
### Spline: Average temperature

- Open the file `in-class-09.Rmd` load the `cities.csv` data and use the subset 
function to keep only the data for the city of Miami.

- We can use the piping operator as well as the subset function to keep 
the data only for the city of Miami

``` r
cities <- cities |> 
  subset(subset = city == "miami")
```

---
### Spline: Average temperature

- Now we can make a box plot of the temperature in Miami by month.

---
### Spline: Average temperature

- Let's start with a linear model that includes 2 nodes and 3 lines.

- We can set a node on the third month and a second node on the 8th month.

- What does this mean? It means that temperature should have a 
consistent rate of change `$(\beta)$` between the months of January to March.

- A different but also consistent rate of change in the months between March 
and August.

- And finally, a different rate of change between the months of August and 
December.

- How can we do this and still maintain the continuity assumption in our model?

- Well we have to use a combination of indicator and difference functions!

---
### Spline: Average temperature

- For example, we know that the average temperature should have an intercept 
different than `$0$` to start.

- Then it might be the case that the temperature increases slightly between the 
months of January to March.

- We can achieve this by combining an indicator function that assigns a value
of 1 to months between January and March and then a 0 to any other month.

- Mathematically we can express this function as:
`$$\mathrm{E}\left(Y_i \mid f_1(\text{month}_i)\right) = \beta_0 + \beta_1 \text{month}_i + \beta_2I(1 \leq \text{month}_i \leq 3) (\text{month}_i)$$`

- This first part of the equation makes sure that the line that connects 
the months of January and March has a single slope and intercept.

---
### Spline: Average temperature

- Add a variable to your data set that assigns the value of 
`$\text{month}_i$` to every observation made in the months of January to 
March and 0 otherwise (question 3 `in-class-09.Rmd`).

---
### Spline: Average temperature

``` r
cities <- cities |> 
  dplyr::mutate("id_jan_mar" = ifelse(test = month <= 3, yes = month, no = 0))
```

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---
### Spline: Average temperature

- With this new function we have that:
`$$\mathrm{E}\left(Y_i \mid f_1(\text{month}_i = 1)\right) = \beta_0 + \beta_1 + \beta_2\\
\mathrm{E}\left(Y_i \mid f_1(\text{month}_i = 2)\right) = \beta_0 + 2\beta_1+ 2\beta_2\\
\mathrm{E}\left(Y_i \mid f_1(\text{month}_i = 3)\right) = \beta_0 + + 3\beta_2 + 3\beta_1$$`

- In this case the parameter `$\beta_1$` represents the rate of change in average
temperature in the city of Miami as a function of the month.

- The parameters `$\beta_1 + \beta_2$` represent the rate of change in temperature
for a month change between the months of January to March. The interpretation of 
`$\beta_0$` depends on how we handle the rest of the lines in the model!

---
### Spline: Average temperature

- Now we have to decide which month will represent the next joint.

- From the data it looks like the trend in average temperature increases at the
same rate in the months between March to August.

- That means that we can have a new line that starts at March and goes all the 
way to August.

- However, remember that we have to satisfy the continuity assumption, which 
means that we want to find a function that makes the following equation true
`$$\lim_{\text{month}_i\to \text{3}^{-}} \mathrm{E}\left(Y_i \mid f_1(\text{month}_i), f_2(\text{month}_i)\right) =\\
\lim_{\text{month}_i\to \text{3}^{+}} \mathrm{E}\left(Y_i \mid f_1(\text{month}_i), f_2(\text{month}_i)\right)$$`

---
### Spline: Average temperature

- Once again let's use an indicator function and a difference function to 
generate our new variable `$f_2(\text{month}_i)$`
`$$f_2(\text{month}_i) = I(3 < \text{month}_i \leq 8)(\text{month}_i - 3)$$`

- This function will satisfy the continuity assumption.

- Add this function as a new variable to your data set with the name 
"id_mar_aug" (question 4 `in-class-09.Rmd`).

- Like before, we can use the `mutate()` function alongside an `ifelse()` 
function to generate our new variable:

``` r
cities <- cities |> 
  dplyr::mutate("id_mar_aug" = ifelse(test = month > 3 & month <= 8,
                                      yes = month - 3, no = 0))
```

---
### Spline: Average temperature

- Finally, we need a third that will allow us to set a different slope 
`$(\beta)$` to the months of September to December.

- Once again we can use an indicator and difference functions in order to make 
sure that we meet the continuity assumption.

- Add a new variable to your data set named "id_aug_dec" that combines an 
indicator function and a difference function to assign values to the months of
September, October, November, and December, that will meet the continuity
assumption (question 5 `in-class-09.Rmd`).

- Like we did before we will use the mutate and ifelse functions:

``` r
cities <- cities |> 
  dplyr::mutate("id_aug_dec" = ifelse(test = month > 8 & month <= 12,
                                      yes = month - 8, no = 0))
```

---
### Spline: Average temperature

- Now we can fit the following model to the data (question 6 `in-class-09.Rmd`):
`$$Y_i = \beta_0 + \beta_1\text{month}_i + \beta_2f_1(\text{month}_i) + \beta_3f_2(\text{month}_i) + \beta_4f_3(\text{month}_i) + \epsilon_i$$`

- Remember that the three functions we have are the same as our new variables
we created, so we can fit this model with

``` r
lm_spline_miami <- lm(
  formula = avgtemp ~ month + id_jan_mar + id_mar_aug + id_aug_dec, 
  data = cities)
```

---
### Plot: expected average temperature

- Make a plot of the average temperature of Miami by month and add the 
estimated expected average temperature from our new model 
(question 7 `in-class-09.Rmd`).

---
### Parameter interpretation

- `$\beta_0$` is the estimated average temperature for the city of Miami at month 
0, like in other models we have used, the intercept is not easy to interpret.

- `$\beta_1$` is the estimated change in the average temperature for the city of 
Miami for a month difference.

- `$\beta_1 + \beta_2$` is the estimated change in the average temperature for 
the city of Miami for a month difference between the months of January to March.

- `$\beta_1 + \beta_3$` is the estimated change in the average temperature for the
city of Miami for a month difference between the months of March to August to December.

- `$\beta_1 + \beta_4$` is the estimated change in the average temperature for the
city of Miami for a month difference between the months of August to December.

---
### Residuals

- Like we have done before, we can look at the residuals of the model to check
if there is a bias in the estimated average temperature.

- Plot the residuals of the model that assumes that the average monthly 
temperature in the city of Miami has a different rate of change between the 
months of January to March, March to August, and August to December 
(question 8 `in-class-09.Rmd`).

---
### The city of San Francisco

- Notice that when we start looking at the data from the city of Miami we made
some assumptions about the position of the joints `$c_1$` and `$c_2$`, however, 
how good would these assumptions be for a different city?

- Load the data for the city of San Francisco in a separate object and add to it
the same 3 variables that we added to the data from Miami 
(question 9 `in-class-09.Rmd`).

``` r
cities <- readr::read_csv(file = here::here("week-05/data/cities.csv")) |> 
  subset(subset = city == "san_francisco") |> 
  dplyr::mutate("id_jan_mar" = ifelse(test = month <= 3,
                                      yes = month, no = 0),
                "id_mar_aug" = ifelse(test = month > 3 & month <= 8,
                                      yes = month - 3, no = 0),
                "id_aug_dec" = ifelse(test = month > 8 & month <= 12,
                                      yes = month - 8, no = 0))
```

---
### The city of San Francisco

- Fit a linear model using these 3 new variables to the average temperature 
of the city of San Francisco (question 10 `in-class-09.Rmd`).

``` r
lm_spline_sfn <- lm(
  formula = avgtemp ~ month + id_jan_mar + id_mar_aug + id_aug_dec,
  data = cities)
```

---
### The city of San Francisco

- Make a plot of the estimated average temperature in the city of San Francisco
as a function of the month according to the linear mode we just fit to the data
(question 11 `in-class-09.Rmd`).

---
### Residuals

- Now that we have applied the same model to both cities we can look at the 
residuals for the city of San Francisco (question 12 `in-class-09.Rmd`).

---
### Linear Splines

- Plotting the estimated expected value of the average temperature of each city 
against the data allowed us to see that

1. The assumptions we made about the city of Miami work relatively well as 
  there is no obvious difference between the line and the median of the data.
  
--

1. The same assumptions applied to the city of San Francisco showed us that 
  a model with more joints could be better, as there are some
  months where the estimated mean is far from the median of the data.
  
--

- It is important to note that for the city of San Francisco we could have found
a better location for the two joints in the model in comparison to using the 
same to points that we used for the city of Miami.

- Those nodes where chosen by looking at data from the city first. We have
no reason to believe that the same joints would work equally well across cities.

---
### Linear Splines

- The problem we would like to solve is: How can we find the best number of 
joints and their position for a given data set?

- There are statistical methods built from the idea of maximizing the 
probability of the data (Maximum Likelihood Estimation), that aim to solve this
problem.

- These method allow us to choose the number and position of the joints for a 
linear spline. Additionally, methods similar to the BIC allow us to penalize 
models that add more joints than what is guaranteed by the data.

- In other words, the arbitrary choices that we made for this example can be 
made using a more systematic approach.

- As a final note, notice that there is no reason why we can't combine the 
splines approach with the non-linear transformations that we used to account
for the average temperature data.